Overwritten virtual does not get called as expected

[Xriuk]

class A{
public:
virtual void func();
};

class B : public A{
void func(bool f = true);
};

A* var = new B();
var->func(); // This should call B's func(), but calls the one from A 

Why is this happening? The default parameter isn't considered as a separated signature (func())?

[ne555]

the signatures don't match, you are not overriding it.
http://en.cppreference.com/w/cpp/language/override

[cire]

The default parameter isn't considered as a separated signature (func())?

Parameters aren't signatures, so your question doesn't make much sense.

But to addresss what I think you're asking, B::func is a function which takes as an argument one value of type bool. A::func is a function which takes no arguments. Providing a default argument for B::func does not in any way change the signature of the function. So, as ne555 says, the signatures do not match.

[TheIdeasMan]

If you use the override keyword where you think you are overriding, the compiler will tell you if you aren't.

class B : public A{
void func(bool f = true) override ;   // <--- error, function doesn't override
};

Good Luck !!

[Xriuk]

You say that I'm not overriding...
But if I call just func() (with no args, and let's say I don't have the default argument in B), then my compiler says that there isn't such function and I need to add "using A::func" to make it work... how is this "behaviour" called?

[keskiverto]

http://stackoverflow.com/questions/411103/function-with-same-name-but-different-signature-in-derived-class